Question

The wires leading to and from a 0.12-mm-diameter lightbulb filament are 1.5 mm in diameter. The wire to the filament carries a current with a current density of 4.3×10⁵ A/m2 .

Part A

What is the current in the filament?

Part B

What is the current density in the filament?

Answer 1

a)

d = 1.5 mm = 0.0015 m

r = d / 2 = 0.0015 / 2 = 0.00075 m

the area of the filament is A = r²

A = 3.14 X 0.00075²

A = 1.76 X 10^-6 m²

but given current density is = 4.3 X 10⁵ A / m²

to calculate the current is = 4.3 X 10⁵ X 1.76 X 10^-6

current i = 0.7568 A

b)
the current density in the filament = current / area

d = 0.12 mm =1.2 X 10^-4 m

r = d / 2 = 1.2 X 10^-4 / 2 = 6 X 10^-5 m

the area of the filament is A = r²

A = 3.14 X (6 X 10^-5)²

A = 1.13 X 10^-8 m²

current density j = 0.7568 / 1.13 X 10^-8

⁼ 66973451.33 A / m²